Question

How do you use a second derivative test where applicable to find all the relative extrema of $f\left( x \right) = {x^2}{\left( {6 - x} \right)^3}$ ?

Answer 1

Hint : Given problem tests the concepts of derivatives and their applications. These type questions can be easily solved if we keep in mind the concepts of maxima and minima. In the problem, we are required to find all the relative extrema of the function $f\left( x \right) = {x^2}{\left( {6 - x} \right)^3}$ given to us. We have to first find the first derivative of the function and equate it to zero so as to find the critical point or the values of x where the slope of tangent of the function is zero. Then, we find the second derivative of the function and confirm whether a point yields a maxima or a minima.

Complete step-by-step answer :
So, we are given a function $f\left( x \right) = {x^2}{\left( {6 - x} \right)^3}$ .
Differentiating the function for the first time using the product rule and chain rule, we get,
$\Rightarrow f'\left( x \right) = {\left( {6 - x} \right)^3}\left( {2x} \right) + 3\left( {{x^2}} \right){\left( {6 - x} \right)^2}\left( { - 1} \right)$
$\Rightarrow f'\left( x \right) = {\left( {6 - x} \right)^3}\left( {2x} \right) - 3\left( {{x^2}} \right){\left( {6 - x} \right)^2}$
$\Rightarrow {\left( {6 - x} \right)^2}\left[ {2x\left( {6 - x} \right) - 3{x^2}} \right] = 0$
$\Rightarrow {\left( {6 - x} \right)^2}\left[ {12x - 5{x^2}} \right] = 0$
Equating the first derivative as zero so as to find the critical points of extremum points, we get,
$\Rightarrow {\left( {6 - x} \right)^2}\left( {12 - 5x} \right)x = 0$
So, we get $x = 0$ , $x = 6$ and $x = \dfrac{{12}}{5}$ .
Now, differentiating the function for the second time, we get,
$f''\left( x \right) = \dfrac{d}{{dx}}\left[ {{{\left( {6 - x} \right)}^2}\left( {12x - 5{x^2}} \right)} \right]$
$\Rightarrow f''\left( x \right) = {\left( {6 - x} \right)^2}\left( {12 - 10x} \right) - 2\left( {12x - 5{x^2}} \right)\left( {6 - x} \right)$
Now, substituting the values of x as $0$ , $\left( {\dfrac{{12}}{5}} \right)$ and $6$ in the second derivative of the function, we get,
So, $f''\left( 0 \right) = {\left( {6 - 0} \right)^2}\left( {12 - 10\left( 0 \right)} \right) - 2\left( {12\left( 0 \right) - 5{{\left( 0 \right)}^2}} \right)\left( {6 - 0} \right)$
$\Rightarrow f''\left( 0 \right) = 36 \times 12 - 2\left( 0 \right)$
$\Rightarrow f''\left( 0 \right) = 432$
Now, $f''\left( 6 \right) = {\left( {6 - 6} \right)^2}\left( {12 - 10\left( 6 \right)} \right) - 2\left( {12\left( 6 \right) - 5{{\left( 6 \right)}^2}} \right)\left( {6 - \left( 6 \right)} \right)$
$\Rightarrow f''\left( 6 \right) = 0$
Also, $f''\left( {\dfrac{{12}}{5}} \right) = {\left( {6 - \dfrac{{12}}{5}} \right)^2}\left( {12 - 10 \times \dfrac{{12}}{5}} \right) - 2\left( {12 \times \dfrac{{12}}{5} - 5{{\left( {\dfrac{{12}}{5}} \right)}^2}} \right)\left( {6 - \dfrac{{12}}{5}} \right)$
$\Rightarrow f''\left( {\dfrac{{12}}{5}} \right) = {\left( {\dfrac{{30 - 12}}{5}} \right)^2}\left( {12 - 24} \right) - 2\left( {\dfrac{{144}}{5} - 5\left( {\dfrac{{144}}{{25}}} \right)} \right)\left( {\dfrac{{30 - 12}}{5}} \right)$
Simplifying further, we get,
$\Rightarrow f''\left( {\dfrac{{12}}{5}} \right) = {\left( {\dfrac{{18}}{5}} \right)^2}\left( { - 12} \right) - 2\left( {\dfrac{{144}}{5} - \dfrac{{144}}{5}} \right)\left( {\dfrac{{18}}{5}} \right)$
$\Rightarrow f''\left( {\dfrac{{12}}{5}} \right) = {\left( {\dfrac{{18}}{5}} \right)^2}\left( { - 12} \right) - 2\left( 0 \right)\left( {\dfrac{{18}}{5}} \right)$
$\Rightarrow f''\left( {\dfrac{{12}}{5}} \right) = - \left( {\dfrac{{324}}{{25}}} \right) \times 12$
So, $x = 0$ is the point of minima of the function $f\left( x \right) = {x^2}{\left( {6 - x} \right)^3}$ as the second derivative of the function for this value of x is positive.
Similarly, $x = \dfrac{{12}}{5}$ is the point of maxima of the function $f\left( x \right) = {x^2}{\left( {6 - x} \right)^3}$ as the second derivative of the function for this value of x is negative.

Note : We can solve the problems involving the maxima and minima concept by two methods: the first derivative test and the second derivative test. The second derivative test involves finding the local extremum points and then finding out which of them is local minima and which is local maxima.